Your suggestion of int[0..1] x f(x) dx suggests you are thinking of
f(x) as a probability density function, which it isn't because the
integral from 0 to 1 of f(x) isn't 1.

Now, *if* f(x) is the cumulative distribution function I propose
above, then its density function would be f'(x) which is 4x^3 on (0,1)
and 0 elsewhere. In that case to calculate the expected value you
would integrate x * 4x^3 = 4x^5.

I'm having a problem finding the mean (U) of this distribution function, f(x) = x to the 4, 0<x<1 How would you solve this problem using the formula:
u=[xf(x)dx] from 0<x<1?

Do you mean f is x cumulative distribution function given by this? :

f(x) = 0 for x <= 0
f(x) = x^4 for 0 < x < 1
f(x) = 1 for x > 1

Your suggestion of int[0..1] x f(x) dx suggests you are thinking of
f(x) as a probability density function, which it isn't because the
integral from 0 to 1 of f(x) isn't 1.

Now, *if* f(x) is the cumulative distribution function I propose
above, then its density function would be f'(x) which is 4x^3 on (0,1)
and 0 elsewhere. In that case to calculate the expected value you
would integrate x * 4x^3 = 4x^5.